In Richard Feynman’s popular book “Surely You’re Joking,
Mr. Feynman!” he tells how during his college years he “often liked to play
tricks on people”. Most of these tricks were designed to show how dumb people
are. For example, in a mechanical drawing class at MIT where the students
were taught to use a drawing instrument called a French curve (a curly piece
of plastic for drawing smooth curves), Feynman informed the other students
that “the French curve is made so that at the lowest point on each curve, no
matter how you turn it, the tangent is horizontal”. He reports that the other
students in the class were excited by this discovery, and began holding up
their French curves and turning them in various ways, trying to verify that
the curve was always horizontal at the lowest point. Feynman found this
funny, because they had already taken calculus and supposedly learned that
the derivative of the minimum of any curve is zero. (Of course, it’s also
intuitively obvious: If a curve at a given point is not flat, the point is
obviously not the minimum.) Feynman says “I don’t know what’s the matter with
people: they don’t learn by understanding; they learn by some other way – by
rote or something. Their knowledge is so fragile!”

A few years later, when he was at Princeton, Feynman
played a similar trick, this time on an assistant of Einstein and hence
presumably well-versed in Einstein’s theory of gravitation. Feynman posed him
a problem:

You
blast off in a rocket which has a clock on board, and there’s a clock on the
ground. The idea is that you have to be back when the clock on the ground
says one hour has passed… Exactly what program of time and speed should you
make so that you get the maximum time on your clock?

According to Feynman, the assistant of Einstein worked on
the problem for quite a bit before realizing that the answer is simply the
natural free ballistic trajectory of an object, because the fundamental
principle of general relativity is that free-fall trajectories (i.e.,
geodesic paths) maximize the elapsed proper time along the path. Feynman
comments that this should have been immediately obvious to Einstein’s
assistant

But
when I put it to him, about a rocket with a clock, he didn’t recognize it. It
was just like the guys in the mechanical drawing class, but this time it
wasn’t dumb freshmen. So this kind of fragility is, in fact, fairly common,
even with more learned people.

Of course, the question was designed to be misleading. As
in the case of Sandford’s elevator,
the misdirection is introduced by referring to the ground, which is actually
irrelevant. If the irrelevant introductory statements had been omitted, and
the question simply posed as “What program of time and speed should you make
so that you maximize the elapsed time on your clock?”, the assistant of
Einstein would presumably have recognized the geodesic condition immediately.

Feynman’s obvious enjoyment of this kind of trick was not
his most loveable quality, although there is undeniably some validity to his
complaint about the fragility of the understanding of many supposedly
intelligent and well educated people. Indeed, someone recently quoted
Feynman’s rocket question (without the punch line) on an internet physics
forum, and participants in the forum began to propose various ways of trying
to solve the problem (unsuccessfully), without realizing that the answer was
simply the geodesic path. My first thought when reading the question was that
it was obviously just describing free trajectories, and I couldn’t believe
that Feynman hadn’t realized this, so I checked the book and found that (sure
enough) he had posed it as a trick question.

Incidentally, the internet poster insisted that it hadn’t
been his intent to repeat Feynman’s trick (perish the thought). Instead, he claimed
to be genuinely interested in the details of how one would go about actually
computing the path of the object that maximizes the elapsed proper time. The
posted responses were all unsatisfactory for numerous reasons. For one thing,
the responses failed to actually connect the metric line element to the
expression to be extremized. For another, the responses took the
Euler-Lagrange equation as given, whereas one would think the derivation of
this equation is precisely what was being sought. Also, the responses tended
to assume the strength of gravity was constant over the trajectory, even
though clearly a one-hour ballistic up-down trajectory would carry an object
to a significant height, where gravity has only a small fraction of the
strength relative to the strength at the surface of the Earth.

The exact solution for radial trajectories is a standard
result in general relativity (see the note on Radial Paths, building on the variational results
from Relatively Straight) but, just for
fun, suppose we were trying to give a simple but complete and self-contained explanation
of how to derive the (approximate) path of maximal proper time to someone who
knows just basic calculus and is willing to accept the Schwarzschild metric line element for
proper time

where m is the mass of the gravitating body (the Earth in
our case), and we are using geometrical units such that G = c = 1. Thus the
differential of proper time along a radial path can be written as

In the weak and slow limit the quantities m/r and (dr/dt)2 are
both extremely small numbers (orders of magnitude smaller than 1), so we can
neglect all products of those two quantities and all higher order terms.
Therefore the spatial curvature part of the metric can be neglected (which would
not be the case if we were considering something involving high speed like
the deflection of a light pulse), and we can take the lowest-order binomial
expansion (1+x)1/2 ≈ 1 + x/2 of the square roots to give the
approximate expression

The elapsed proper time Dt along a path r(t) from t1 to t2
is the integral of this, i.e.,

We want to find the function r(t) that maximizes this elapsed proper time. To
understand conceptually how we do this, it may be helpful to recall how we
find the maximum of an ordinary function f(x). We find x such that df/dx
= 0. Strictly speaking, this just locates a point where the function is
stationary, i.e., flat (like the bottom of Feynman’s French curve), so it can
represent either a maximum, a minimum, or an inflection point. But the
context usually makes it clear which of these applies.

Now, to find the function r(t) that maximizes the integrated Dt, we proceed in a similar way, by finding a path for which Dt is stationary for small variations in the path. For any
small constant u and any arbitrary function q(t) defined from t1
to t2 with q(t1) = q(t2) = 0, let Dt(u) denote the integrated proper time for the disturbed path
r(t) + uq(t). Thus we have

The disturbed path represented by r(t) + uq(t) has the same starting and
ending points as r(t), but deviates from r(t) between those points. Our
objective is to find a function r(t) such that d(Dt)/du = 0
at u = 0 for any q(t). This condition signifies that Dt is stationary under any small variations of the path.

To differentiate the integral for Dt(u)
with respect to u we can bring the differentiation inside the integral, so we
have

where accent symbols signify derivatives with respect to
t. At u = 0 this reduces to

Now, recalling from basic calculus that (r'q)' = r'q' + r''q, we can write this
as

The integral of the middle term from t1 to t2 is simply
r’(t2)q(t2) – r’(t1)q(t1), which
equals zero because q(t) is zero at both t1 and t2. Hence
that term drops out, and we have

To make r(t) a stationary path, this derivative must equal zero for all possible
deviation functions q(t), which implies that the expression in the square
brackets must vanish. Hence we have the condition for the path length to
be stationary

This is the geodesic equation for radial trajectories (to the lowest order of
approximation), and of course it is just Newton's inverse square law of
gravitation combined with the first law of motion, confirming that general
relativity does indeed reduce to Newtonian gravity in the weak slow limit.
(Coincidentally, if we replace the coordinate time t with the proper time t, this formula is also exact in general
relativity for a momentarily stationary free test particle, as explained in Radial Paths.)

Our next task is to solve this equation for the actual path. This too is
a standard result in classical mechanics. (See for example the note on Free-Fall Equations.) Integrating both
sides of the geodesic equation over dr from an arbitrary initial separation
r(0) to the separation r(t) at some other time t gives

The left hand integral can be rewritten as

Therefore, the previous equation can easily be integrated
to give

Solving the equation for r’(t), we have

Rearranging this gives

To simplify the expressions, we put r0 = r(0), v0
= r’(0), r = r(t), and r = r/r0.
In these terms, the preceding expression can be written

If K is positive the trajectory is bounded, and there is
some point on the trajectory (the apogee) at which v = 0. Choosing this point
as our time origin t = 0, we have K=1, and the standard integral gives

It’s useful to define a parameter q such that

Inserting this expression for r into the preceding
equation, we find that t can be expressed in terms of the parameter q by

These two parametric equations describe a cycloid, which
is the curve traced by a point fixed on the perimeter of a wheel rolling
along a flat surface, as illustrated in the figure below.

Our final task is to insert the specific conditions of the
problem, i.e., a one hour radial round-trip beginning and ending at the
surface of the Earth at radius re, to determine the actual the
height r0 of the apogee where q
= 0 and t = 0, and the cycloid parameter q*
where r = re. We know q =
0 and t = 0 at the apogee, where r = r0, and we also know r equals
the radius of the Earth re at the times t = ±1800 seconds. (It
takes half an hour to reach the apogee, and another half hour to fall back to
Earth). We need to determine the value of q
corresponding to the beginning and ending of the trip.

From the radial parametric equation we have

Solving this for r0 and substituting into the
temporal parametric equation gives

The mass of the Earth in geometric units is m = 0.00443
meters, the radius of the Earth is re = 6378000 meters, and the
value of t* in geometric units is (5.4)1011 meters
(which is half an hour times the speed of light). Inserting these values, we
have

which can be solved numerically to give q* = 1.494. The radial
parametric equation then gives the radial position of the apogee r0
= 11850000 meters. (Since this is nearly twice the radius of the Earth, the
acceleration of gravity at that height is only 1/4 as great as at the Earth’s
surface, which confirms that we cannot assume constant gravity.) A plot of
the relevant portion of the cycloid is shown below.

Also from the ratio of dr/dq
and dt/dq the velocity is given by

and so the velocity of the projectile at the Earth’s
surface is 7596 meters/second, which is about 68% of escape velocity. These
parametric equations represent the “program of time and speed so that you get
the maximum time on your clock”. To give the speed as an explicit function of
radial position, we can solve the radial parametric equation for q and substitute into the speed equation,
giving the familiar result

Since we’ve been working only to the lowest order of
approximation, it was permissible to evaluate the trajectory for a one-hour
lapse of proper time for the ballistic object, because the difference between
that and the proper time for a stationary clock on the Earth’s surface is
extremely small. To evaluate this difference, we can use the exact expression
for the elapsed coordinate time derived in the note on Radial Paths, and then multiply this by dt/dt (also derived in that note) for the Earth’s
surface to give the trip time as measured by the ground clock. For a round
trip that takes 3600 seconds (1 hour) on the ground clock, the elapsed proper
time on the ballistic clock exceeds this by 1.772 msec, and the elapsed coordinate time exceeds the ground
clock by 3.655 msec.